# What is the condition on the sequence for this?

$\left[ \frac{1}{2}+\frac{1}{\pi}\tan^{-1}(\lambda_n)\right]^n\rightarrow 1$.

I want to find the condition on $\lambda_n$ which leads the above display as $n$ tends to $\infty$.

I guess that the condition would be $\lambda_n \succ n$, but I cannot show it in a rigorous way.

• Could you explain what connection this question has with statistics? – whuber Nov 12 '15 at 0:04
• This comes from a statistical problem. For example, one can show that $P\{\sum_{i=1}^n \exp( W_i/2 ) >n \lambda_n \}\leq 1- [P(Q \leq \lambda)]^n$, where $W_i$ is from i.i.d $\chi^2_1$ for $i=1,\cdots,n$ and $Q$ follows Cauchy distribution. Also, $[P(Q \leq \lambda)]^n = [1/2+\tan^{-1}(\lambda_n)/\pi]^n$. When $W_i$ is a test statistic, it is related to a multiple testing problem. – Minsuk Shin Nov 12 '15 at 2:19
• It's a typo; $P(Q\leq \lambda)$ should be $P(Q\leq \lambda_n)$. – Minsuk Shin Nov 12 '15 at 2:31

If $\left[X_n\right]^n \rightarrow 1$ as $n \rightarrow \infty$, then $X_n \rightarrow 1$ as $n \rightarrow \infty$.
It follows that $\left[\frac{1}{2} + \frac{1}{\pi}\tan^{-1}\left(\lambda_n\right)\right]$ must tend to 1 as $n$ tends to $\infty$.
Therefore, $\frac{1}{\pi}\tan^{-1}\left(\lambda_n\right) \rightarrow \frac{1}{2}$, which implies that $\tan^{-1}\left(\lambda_n\right) \rightarrow \frac{\pi}{2}$.
Finally, this requires that as $n \rightarrow \infty$, $\lambda_n$ must tend towards $-\tan\left(\frac{\pi}{2}\right)$
• Thankyou for the comments. This is obvious in a sense that $\tan(\pi/2)=\infty$ and implicitly it is assumed that $\lambda_n \rightarrow \infty$. I am interested the rate of $\lambda$. – Minsuk Shin Nov 12 '15 at 2:29